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For sochastic processes $X_n$ and $Y_n$ it holds that if they are indenpendant for all $n$ then $$E[X_nY_n] = E[X_n]E[Y_n]$$ But can you go the other way. I mean if $E[X_nY_n] = E[X_n]E[Y_n]$, does that mean that $X_n \text{ and }Y_n$ HAVE to be indenpendant. Basically what I am asking is which of these following statements is true:

Statement 1: $X_n$ and $Y_n$ are indenpendant $\Rightarrow $ $E[X_nY_n] = E[X_n]E[Y_n]$

Statement 2: $X_n$ and $Y_n$ are indenpendant $\Leftrightarrow $ $E[X_nY_n] = E[X_n]E[Y_n]$

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Statement 1 is true if the expectations exist; statement 2 is not necessarily true.

Assuming $X_n$ and $Y_n$ have finite mean and variance then $E[X_nY_n] = E[X_n]E[Y_n]$ is equivalent to saying they have zero covariance and correlation. But this does not mean they are independent.

As a counterexample, consider $X_n$ having a standard normal distribution $N(0,1)$, and independently $Z_n$ taking the values $-1$ and $1$ with equal probability, and $Y_n = X_n Z_n$.

Then $E[X_n]=0$, $E[Y_n] = E[X_n]E[Z_n]=0$, and $E[X_nY_n] = E[X_n^2]E[Z_n]=0$ so $E[X_nY_n] = E[X_n]E[Y_n]$

but (apart from sign) $X_n$ determines $Y_n$ and they are not independent

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  • $\begingroup$ This is not quite right. $E[Z_n]=1/2$ $\endgroup$ Commented Jan 28, 2018 at 3:32
  • $\begingroup$ The modified example works. i had no doubt that you could create a counterexample. It is just the one you chose didn't work. $\endgroup$ Commented Jan 28, 2018 at 11:43

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